Pressure Challenge Question

For Q1=0, it is readily appearant that P1=P2=P3. Since we are neglecting losses, this holds true for all cases. This is true even if Q2 is not equal to Q2 because of piping arrangements if pressure losses due to flow are ignored.

Your question states that, "....for a constant flow rate..."...

As Guest above said neglecting losses the pressures will all be equal...

However, the flowrate will be halved in the split pipes (if everything is equal) as the velocity is halved.

John.

Corrrection from my original challenge question:

The constant flow rate refers to the original pipe. There is a constant flow rate in the pipe before it splits, and there is no change in flow with respect to time.

As there is no loss and no change in elevation the power flowing through pipe 1 section is double that through a section through pipe 2 or 3. As velocity of flow in pipes 2 & 3 needs to be half that in pipe 1 ( for continuity) the total kinetic energy will be only a quarter of that in pipe 1. So to keep the hydraulic head same the pressure has to go up. So P2 = P3 and P1 < P2 & P3.

Agreed. Wikipedia (as ever) provides 'chapter and verse' in the sections on Bernoulli's equation and Bernoulli's principle. We can also think of this arrangement as a special case of the Venturi effect.

I agree with Yanthram and Physicst and here is the proof

Since the two branches are identical to the inlet pipe then the flow rates in them Q₂ and Q₃ must also be equal and therefore half the flow rate Q₁. And we can also state that the velocity of the fluid in the supply pipe V₁ is twice the velocity V₂ of the outlet pipes.Bernoulli's states that due to the conservation of energy

Where

K = arbitrary constant for the system

v = Velocity

g = Acceleration due to gravity

h = height above and arbitrary level

p = pressure

ρ = density of the fluid

so for our system then

Therefore we can see that since V₁² must always be positive then P₂ must always be greater than p₁ while the velocity in greater than 0

I'm not sure that bernoulli can be applied from 1 to 2 as you have shown. Since point 2 does not encompass all the fluid flow that originated in point 1. I have run the numbers and performed a bernoulli from 1 to 2 already. It shows a pressure increase as does your proof. But all of the parallel pipe examples that I can find in my old text books run bernoulli from 1 to 4 (4 is the theoretical point where the two channels merge back to one). I agree with your proof of bernoulli, but I think it is not applicable to this problem due to the fact that you do not have all the flow in section 2 that you have in section 1.

Bernoulli is never wrong...but if you give bernoulli junk, he will give you nothing

Think of it like this. If it were flowing from one pipe with a cross sectional area of a to a second pipe of cross sectional area 2a then Bernoulli's principal would apply. Since the only thing we have changed from the above scenario is to use two pipes with cross sectional area a rather than one with 2a then it still applies. The only difference would be the resistance to the flow caused by the friction on the walls of the pipe as the twin pipe system would have proportionally more surface area to volume than the single pipe.

I expect the first comment below is obvious in the context of what Masu said, but:
As there are stated to be no losses we can generalise much further - that the pressure-velocity relationship holds independently of the number and relative dimensions of the pipes in which the velocity is induced. The following thought experiment supports this: divide the pipe into two, maintaining constant total area - therefore no pressure change. Now we can vary the areas of the pipes independently to give independent flow rates - and the equations clearly continue to apply.
This leads us to a paradox:
When we perform an experiment to demonstrate Bernoulli's equation we use stationary liquid to monitor the pressure of the flowing liquid. So why isn't the pressure of the stationary liquid always the same?
Enjoy...

stationary liquid pressure will be the same unless there is an elevation or diameter difference

You can't neglect any loss or affect/effect. All fluid in motion systems are dynamic and directly a result of thier added losses. The smaller pipes can't dynamically be equal if they are of 1/2 the area of the larger pipe. Because the ratio of pipe inside surface to pipe area affect the drag losses differently in the two different pipe diameters. What you are describing will not exist in any prectical application. The smaller pipes should be adjusted up in area to overcome the effects of higher loss ratios.

Guest the question states that

"the diameter of both branches are equal to each other as well as equal to the original pipe"

Which to my way of thinking means they are all the same diameter.

correct

The pipes are all stated to be the same diameter - i.e. the output pipe area is greater than the input pipe. And... perhaps you never need to isolate effects temporarily (as implied in this question) to develop understanding? So I guess you are far too clever to be an engineer

Bernoulli's equation applies along a streamline. So Masu's analysis is valid. To better perceive this problem, think of the total pressure (pressure head + velocity head) remaining constant in absence of losses. Or consider it this way: A force must cause a fluid to accelerate/decelerate - declerate in this case into the two pipes. No force is provided by friction or gravity in this case, so the force is provided by the pressure difference.